A real valued function f(x) satisfies the functional equation f(x – y) = f(x) f(y) – f(a – x) f(a + y) where a is a given constant and f(0) = 1, f(2a – x) is equal to
–f(x)
f(x)
f(a) + f(a – x)
f(-x)
A.
–f(x)
f(a – (x – a)) = f(a) f(x – a) – f(0) f(x)
= - f(x) [ ∵ x = 0, y= 0, f(0) = f^{2} (0)-f^{2}(a) = 0 ⇒ f(a) = 0]
Suppose f(x) is differentiable x = 1 and
3
4
5
6
C.
5
Let f and g be differentiable functions satisfying g′(a) = 2, g(a) = b and fog = I (identity function). Then f ′(b) is equal to
1/2
2
2/3
None of these
A.
1/2
Given fog = I
⇒ fog(x) = x for all x
⇒ f ′(g(x)) g′(x) = 1 for all x
f ′(g(a)) =1/g('a) = 1/2
⇒f(b') = 1/2
The range of the function ^{7-x}P_{x-3} is
{1, 2, 3}
{1, 2, 3, 4, 5}
{1, 2, 3, 4}
{1, 2, 3, 4, 5, 6}
A.
{1, 2, 3}
The given function f(x) = ^{7-x}P_{x-3} would be defined if
(i) 7 - x > 0 ⇒ x < 7
(ii) x - 3 > 0 ⇒ x > 3
(iii) (x - 3) < (7 - x)
⇒ 2x < 10 ⇒ x < 5
⇒ x = 3, 4, 5
Hence Range of f(x) = {^{4}P_{0}, ^{3}P_{1}, ^{2}P_{2}}
Range of f(x) = {1, 3, 2}
Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)} be a relation on the set A = {3, 6, 9, 12} be a relation on the set A = {3, 6, 9, 12}. The relation is
reflexive and transitive only
reflexive only
an equivalence relation
reflexive and symmetric only
A.
reflexive and transitive only
Reflexive and transitive only.
e.g. (3, 3), (6, 6), (9, 9), (12, 12) [Reflexive]
(3, 6), (6, 12), (3, 12) [Transitive].
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is
a function
reflexive
not symmetric
transitive
C.
not symmetric
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} is a relation on the set A = {1, 2, 3, 4}, then
(a) Since ∈ R and (2,3) ∈ R, so R is not a function.
(b) Since (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉ R. So R is not transitive.
(c) Since (2, 3) ∈ R but (3, 2) ∉ R, so R is not symmetric.
(d) Since (4, 4) ∉ R so R is not reflexive,
A function is matched below against an interval where it is supposed to be increasing. Which of the following pairs is incorrectly matched?
Interval | Function |
(-∞, ∞) | x^{3} – 3x^{2} + 3x + 3 |
Interval | Function |
[2, ∞) | 2x^{3} – 3x^{2} – 12x + 6 |
Interval | Function |
(-∞, 1/3] | 3x^{2} – 2x + 1 |
Interval | Function |
(- ∞, -4] | x^{3} + 6x^{2} + 6 |
C.
Interval | Function |
(-∞, 1/3] | 3x^{2} – 2x + 1 |
Clearly function f(x) = 3x2 – 2x + 1 is increasing when
f′(x) = 6x – 2 ≥ 0 ⇒ x∈[1/3, ∞)
Let f : (-1, 1) → B, be a function defined by then f is both one-one and onto when B is the interval
[0, π/2)
C.
The domain of the function
[2, 3]
[2, 3)
[1, 2]
[1, 2)
B.
[2, 3)
For the the function
we will define
(I) −1≤ (x − 3) ≤1⇒ 2 ≤ x ≤ 4 ....... (i)
(II) 9 x 0 3 x 4 2 − ≥ ⇒ − < < ....... (ii)
From relation (i) and (ii), we get 2 < x < 3
∴ Domain of the given function = [2, 3)
The graph of the function y = f(x) is symmetrical about the line x = 2, then
f(x + 2)= f(x – 2)
f(2 + x) = f(2 – x)
f(x) = f(-x)
f(x) = - f(-x)
B.
f(2 + x) = f(2 – x)
If y = f (x) is symmetric about the line x = 2 then f(2 + x) = f(2 – x).